#### Answer

$$
4 y^{\prime \prime}-4 y^{\prime}-3 y=0
$$
The general solution of that equation is given by
$$ y=c_{1} e^{\frac{3}{2}t} +c_{2} e^{ -\frac{1}{2}t} $$
where $ c_{1} $ and $c_{2}$ are arbitrary constants.

#### Work Step by Step

$$
4 y^{\prime \prime}-4 y^{\prime}-3 y=0 \quad \quad (1)
$$
We assume that $ y = e^{rt}$, and it then follows that $r$ must be a root of the characteristic equation
$$
4r^{2}-4 r-3=(r+\frac{1}{3})^{2}=(r-\frac{3}{2})(r+\frac{1}{2})=0,
$$
Thus the possible values of $r$ are $r_{1}=\frac{3}{2} , r_{2}=-\frac{1}{2}$; the general solution of Eq. (1) is
$$ y=c_{1} e^{\frac{3}{2}t} +c_{2} e^{ -\frac{1}{2}t} $$
where $ c_{1} $ and $c_{2}$ are arbitrary constants.